Consider a block kept on an inclined plane (inclined at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane ($$\mu$$) is equal to: 

Let mass $$=m,angle, θ=45°$$

Normal reaction:
$$N=mg\cosθ$$

Friction:
$$f=μN=μmg\cosθ$$

Now case 1: just moving up the incline

Friction acts down the plane

$$F₁=mg\sinθ+μmg\cosθ$$

Now case 2: just preventing sliding down

Friction acts up the plane

$$F₂=mg\sinθ−μmg\cosθ.$$

Given:

$$F_1=2F_2$$

Substitute:

$$mg\sinθ+μmg\cosθ=2(mg\sinθ−μmg\cosθ)$$

Cancel mg:

$$\sinθ+μ\cosθ=2\sinθ−2μ\cosθ$$

Bring terms together:

$$\sinθ−2\sinθ=−2μ\cosθ−μ\cosθ$$
$$−\sinθ=−3μ\cosθ$$

So:

$$μ=\frac{(\sinθ)}{(3\cosθ)}$$
$$μ=\frac{(\tanθ)}{3}$$

Now$$θ=45°,\tan45=1$$

So:

$$μ=\frac{1}{3}$$

Final answer:

$$\frac{1}{3}$$